Quotient maps: If $X$ is locally convex space, then so is $X/N$. So I am having trouble understanding quotient map in my functional analysis class. We are given the following exercise for practice. let $X$ be a topological  vector space and let $N\subseteq X$ be a closed subspace. Take $Q:X\rightarrow X/N$ to be the quotient map. I've managed to show that if $p$ is a semi-norm on $X$, then $p_N([x])=inf\{p(x+y):y\in N\}$ is a semi-norm on $X/N$. The next step is to show that the $Q(\{y\in X:p(x-y)<r\})=\{[y]\in X/N:p_N([x]-[y])<r\}$, which I am having issues with. I just can't get my head around what the LHS looks like. The next part of the exercise is to show that the space $X$ being a locally convex space implies that $X/N$ is as well. Any help with that part would be appreciated as well. Thanks in advance.
 A: For the first part, wlog $x=0$, and let $B_r=\{y\in X:p(y)<r\}$ and $B_r^N=\{[y]\in X/N:p_N([y])<r\}$.
If $y\in B_r$ then by definition of $p_N$, we have $p_N([y])\le p(y)<r$ so $ Q(B_r)\subseteq B^N_r.$ 
On the other hand, if $[y]\in B_r^N$, then choose $n\in N$ such that $p(y+n)<r$. Then, $y+N=y+n+N=Q(y+n)\Rightarrow B_r^N\subseteq Q(B_r).$ The result follows. 
For the second part, let $Q : X → X /N$ be the quotient map. Choose a neighborhood $V$ of $0$ in
$X /N$. Since $V$ is a neighborhood of $0$ and $Q$ is continuous, the pre-image $Q^{-1}(V)$ is a neighborhood of $0$ in $X$. Now, since $X$ is locally convex, there exists an open convex set $U \subseteq X$ , with $0 \in U \subseteq Q^{-1}(V)$. Then, $Q(U)\subseteq V$ is open (since $Q$ is open). The fact that $Q(U)$ is also convex is almost immediate:
suppose $[x], [y]\in Q(U)$ and consider $t[x]+(1-t)[y];\ 0\le t\le 1$. Expanding, we get $t(x+N)+(1-t)y+N=tx+(1-t)y+N$ and now the result follows since $tx+(1-t)y\in U$ (since $U$ is convex), which means that $t[x] +(1-t)[y]=\in Q(U).$
